Let the magnitude of the velocity at which the ball is thrown be V and the angle with the horizontal it is thrown at equal X.

The vertical component of the velocity of the ball is given by V*sin X and the horizontal component of the velocity of of the ball is V*cos X.

As the vertical component of the ball's velocity is given as 0.5 m/s and the horizontal component is given as 3.6 m/s, the magnitude of the velocity it is throw at is sqrt(0.5^2 + 3.6^2) = sqrt(0.25 + 12.96) = sqrt(13.21) m/s

The angle at which the ball is thrown is equal to an angle of arc tan (0.5/3.6) = 7.907 degrees with the horizontal.